Question

The speed of a uniform solid sphere after rolling down from rest without slipping along a fixed inclined plane of vertical height ' $h$ ' is______

• A. $\sqrt{\frac{10gh}{7}}$
• B. $\sqrt{gh}$
• C. $\sqrt{\frac{6gh}{5}}$
• D. $\sqrt{\frac{4gh}{3}}$

Answer 1

Answer: (A)

If $v$ is the speed of solid sphere when it reaches at point $O$, then by the law of conservation of energy,

$mgh=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$
$=\frac{1}{2}\cdot \frac{2}{5}m{R}^2{\left(\frac{v}{R}\right)}^2+\frac{1}{2}m{v}^2$
$\Rightarrow mgh=\frac{1}{5}m{v}^2+\frac{1}{2}m{v}^2$
$\Rightarrow gh=\frac{v^2}{5}+\frac{v^2}{2}$
$\Rightarrow gh=\frac{7v^2}{10}$
$\Rightarrow v^2=\frac{10gh}{7}$
$\Rightarrow v=\sqrt{\frac{10gh}{7}}$